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Theorem
8
the
divergence
theor em
Gauss
's
Theorem
)
Let
D
be
a
regular
.
3-
dim
.
domain
whole
boundary
-5
is
an
oriented
,
closed
surface
with
unit
normal
field
Ñ
pointing
out
of
D.
Recall
the
definition
of
flux
of
aviator
field
4-
=P
is
a
smooth
Vertov
field
defined
on
D
,
Given
any
continuous
vector
field
=P
.
the
flux
of
É
then
across
the
oriented
surface
5
is
the
integr al
fffdiv
F-
du
=
☒
I.
Ñds
.
of
the
normal
component
of
=P
over
S
,
D
5
If
É
.
Ñds
=
If
E.
D8
Sol
.
Let
D
be
the
solid
ball
given
5
5
by
txt
)
't
+
1)
7-
(2--2)
'
E
4
.
When
the
surf ace
is
closed
,
the
flux
intesral
can
be
Then
D
is
a
regular
.
3-
dim
domain
whose
devot ed
by
☒
E.
ÑDS
=
☒
E.
D8
.
boundary
S
is
the
sphere
I
5
☒
-
1)
4-
+
1)
2+(2--2)
'
=
4
.
Computation
method
.
which
is
an
oriented
.
closed
surface
,
with
1.
By
definition
.
unit
normal
vector
field
Ñ
pointing
out
9-
D.
2.
The
divergence
theorem
.
(
Gauss
's
Theorem
)
wanminliu@gmail.com
Method
1
we
change
variables
.
Let
The
vector
field
-7=1×2
,
y
?
2-4
is
U
=
X
-
l
,
V
=
y
-11
,
smooth
over
D
.
W
=
2-
-2
,
then
du
=
dx
,
dv
=
dy
,
dw
=
dz
Then
by
the
Gauss
's
Theorem
.
the
flux
dV=
dxdydz
=
dudvdw
has
the
expression
21×+7+2-1=2
(
U
-11-1
V
-
I
+
W
-12
)
=
zlutvtw
)
+
4
Flux
-
-
☒
E.
Ñds
=
fff
dived
V
⑧
5
☐
⑧
=
US
21×+7+2
)dY
=/
fffzlutvtw
)
+
4)
dudv dw
D
u4v4wk4
dive
=P
.
É
=
1¥
.
¥
,
1.
(
x
!y
!
E)
Since
U
is
an
odd
function
on
the
symmetric
domain
u2+v4w
's
4
,
we
hav e
=
1*1×4
+
b)
t
12-4
ffs
u
dudvdw
=
0
.
u¥v4w%4
Similarly
,
SIS
zlutvtw
)
dadvdw
=
0
.
=
21×+7-1
Z
)
.
u4vFwkx
We
have
two
methods
for
the
computation
.
Then
aux
=
fff
4
dudvdw
=
4
Voll
Ball
)
ñ+i-wkx_
Ballot
radius
2
=
4
.
-452-23=13-284
☐
wanminliu@gmail.com
Method
.
Then
fff
div
ÉDV
I
=
1
.
5=-1
.
I
=
2
D
=
zfffcxtytt
)dV
So
D
fff
dw
DX
=
211-1+2
)
VID
)
D
=
4
VID
)
=
2
⑦
+
y_
+
E)
VCD
)
=
4.
a
-2
>
=
a.
.
ISS
✗
DX
Here
☒
=
1¥
VCD
)=vd
(D)
☐
i=¥÷
2-
=
VCD
)
are
centroid
of
D.
since
D
is
a
ball
with
center
CI
.
-1.2
)
wanminliu@gmail.com
rot
=P
=
curl
¥
is
avatar
field
.
So
by
the
Gauss
's
Theorem
(
divergence
Theorem
!
f)
Gave
•
Ñds
=
V1
div
Karl
IV.
SUA
D
Notation
.
rot
=
curl
É
=p
✗
§
union
ots
and
A
Sol
.
(
method
1
.
Divergence
Theor em
)
div
(
care
E)
Let
D
be
the
region
of
half
=
dir
(
☐
✗
E)
solid
ball
✗
1972-4-4
and
2-70
.
=p
.
(0×-7)
with
an
outward
pointing
unit
normal
field
=
0
(
Theorem
3
(g)
)
Ñ
.
Then
boundary
of
D
consists
of
S
So
ff
full
Nils
+
Island
É
)
-
ÑadS=ffodV=0
.
and
the
disc
A-
=/
IX.
4,0
)
/
✗
Fyi
4
}
S
A
D
The
unit
normal
vector
field
→
y
for
points
on
A
is
Have
E)
Ñsds
=
-
Have
-51
-
Ñads
s
A
£
¥
A
Ña=(
o.o
,
-1
)
.
(
outward
D)
wanminliu@gmail.com
JE
Method
2
by
Stokes
's
Theorem
.
cure
=/
¥×¥¥
A-
y
y
,
*HÉtyJ_
I
let
5
be
a
piecewise
smooth
.
oriented
surface
in
3-
space
,
with
unit
normal
field
Ñ
and
boundary
e
cousin
Ting
of
one
or
more
piecewise
curl
É
.
ÑA
=
(1×-7)
,
-7
1)
•
I
0
.
°
.
-4
gmoth
closed
came
,
with
orientation
inherited
from
5
.
If
É
is
a
smooth
Vertov
field
=
-1
.
defined
on
an
open
set
containing
5
.
Then
so
§
cure
É
)
-
Ñsds
ff
care
E.
Ñds
=
§
F.
dñ
s
e
=
-
IS
find
É
)
.
Ñn
.
Is
=
If
ds
A
A
=
Area
(
A
)
=
I.
22
=
47
.
4
radius
2
☐
wanminliu@gmail.com
(
method
2
,
Stokes
's
Theorem
)
so
§
awe
.
Ñds
Sol
.
Let
e
be
the
circle
=
§
É
.
dÑ
=
§
(
3×-7
.
YZ
,
✗
7)
(
dx.dk
°
)
it
{
IX.
y
,
031×45=22
}
e
e
¥:
>
y
with
orientation
in
the
picture
=
§
(3×-4)
dxtyz
dy
E
-
q
e
2-
=
o
one
Then
ff
cure
.
ÑdS=§É
.
tar
=
§
(3×-9)
dx
s
e
E
✗
=
2
Cost
with
0
C-
[
0.2-a
)
d.
f
=
(
dx.dy.dz
)
=
(
IX.
DY
.
°
)
y
=
25in
0
So
Ix
=
245m03
do
Moreover
,
for
point
ix.
yo
)
on
the
circle
|
=
/
ok
¢
wso
-
zsinofzsino
)
do
✗
7-
y
'
=
22
we
have
=
4
so
"si÷
do
-
6)
Find
do
Zxdx
+
ZYIY
=
0
.
⇐
-
game
,
25140000W
=
4K
.
0
.
☐
wanminliu@gmail.com
§
17.3
Proof
.
Let
Brca
.
b)
be
the
disk
#
9
(
Average
values
of
harmonic
functions
of
radius
r
with
center
I
a.
b)
in
IR
?
If
UC X
,
y
)
is
a
harmonic
function
in
suppose
that
Brea
.
b)
is
contained
in
a
domain
containing
a
disk
of
radius
r
the
domain
of
UCX
,
y
)
.
Recall
that
UC X
.
y
)
is
called
harmonic
with
boundary
Cr
,
then
the
average
if
0
Ulx
,
y
)
=
0
,
val ue
of
u
around
the
circle
is
the
val ue
of
u
at
the
center
.
With
D=
0.0
=
¥
.
+
Ep
.
Prove
this
by
showing
that
we
want
to
show
that
•
the
derivative
of
the
average
value
Ula
.
b)
=
¥
§
UCX
in
ds
with
respect
to
r
is
zero
using
the
2Br(
a.
b)
Diverge nce
Theorem
and
harmonic
"7°t
4
Define
g-
(
r
)
=
¥
,
§
UH
,
Dds
.
•
him
of
the
average
values
as
r
-70
2
Bria
.
b)
is
the
value
of
u
at
the
center
.
We
wan t
to
show
that
f-
'
(
r
)
=
0
.
wanminliu@gmail.com
We
parametrize
circle
Crl
a.
b)
with
Take
É=
☐
u
=
l¥u
.
2-
u
)
NO
)=
(
✗
101
,
YIO
)
)
i
Then
7.
É
=
V.
ou
=
"
+
÷
"
q
o
(
a.
b)
✗
(
O
)
=
At
r
↳
0
oggyzz
,
go
since
U
is
harmonic
.
YLO
)
=
btrstno
D=
f)
☐
•
(
Ux
.
Uyjdxdy
=
fux
dy
-
Uyetx
g-
'
(g)
=
f-
r
stall
,
r
↳
0
)
Blais
)
erla.hn
>
✗
Is
=
rdo
=
[
"
Uxfatrcoio
,
btrino
)
rooiodo
2k
f-
(
r
)
=
¥
,
§
UH
.Dds=÷fou(
atrwso.b-rsinordll-fokuyfa-rooo.b-rstnolrstnod02Brla.to
)
*
chain
rule
=
¥
/
Fula
+
rwso
.
btrstnos
do
=/
ok
(
Ulatrcoio
,
btrstnoi
)
do
Leibniz
rule
we
want
to
show
firs
=
0
.
=
¥r(
So
"
ulatrwso.btrs.no
)
do
)
*
Then
fer
]
=
constant
Recall
that
for
=T=
(
Fixx
.
⇐
ix.
y
,
)
,
=
(
zafer
)
)
f)
☐
•
Édxdy
=
§
F
,
dy
-
Fzdx
on
the
other
hand
,
Brlab
)
e
him
fer
,
=
u(
a.
b)
by
mean
value
theorem
and
continuity
of
UCX
.
4)
r→o
[
This
is
a
variation
of
"
Theorem
6
"
in
textbook
)
go
the
constant
is
Ula
.
b)
and
fer
)=
Ula
.
b)
☐
wanminliu@gmail.com
§
/
7
.
5
Proof
.
#
5
Use
Stokes
's
Theorem
to
show
that
let
=
(
y
,
z
,
x
)
.
Since
dk-ldx.dy.dz
)
.
§
ydxttdytxdz
=
Ex
a
'
we
hav e
e
§
ydxtzdy
+
Xdz
=
§
É
.
di
e
e
where
e
is
the
suitably
oriented
intonation
of
sulfa
Is
x4y4
2-
'
=
a
'
=
f)
car e
É
.
Ñds
and
x
-17+7=0
.
5
Siaa
the
plane
10.0.14
?
¥pÑ
✗
+7+2=0
passes
through
the
origin
0
.
Recall
§
E.
di
=
ff
cure
É
.
Ñds
5
which
is
the
center
of
spher e
,
e
s
B
so
the
plane
out
the
sphere
into
two
semis
phone
.
I
oriented
surface
with
unit
normal
Ñ
We
pick
up
S
as
the
e
boundary
of
5
with
inherited
orientation
femispbe.ve
which
contains
10,0
,
lad
,
with
outward
orientation
,
and
boundary
off
is
e.
>
<
UÑ
wanminliu@gmail.com
Then
boundary
of
D=
S
U
B
iii.
curl
=/
¥
¥
¥
/
=
Hit
,
-11
with
orientation
of
B
✗
a-
×
B
.
n=
Let
P
=
IX.
Y
,
Z
)
be
a
point
on
Gani
,
;
div ergence
theorem
f
the
sphere
✗
4-
y
't
2-
'
=
a-
Have
Ñdstffame
E.
ÑdS=,§d÷ÉtdV=°
.
5
B
Then
Ñp
=
=
so
Youre
E.
Ñds=
-
Heure
F.
Ñds
B
(
1.1.11
cure
É
.
Ñp
=
Ya
,
C-
✗
-
y
-
Z
)
=
-
ISH
.
-1
.
-
D.
'
ds
!¥
B
=
-53%
IS
=
-
Baa
'
Let
B
be
the
disc
on
the
plane
✗
+
y
-12=0
with
xfyfzla.az
,
To
kill
the
negative
sign
,
we
need
to
take
the
orientating
e
to
be
clockwise
and
D
be
the
half
ball
of
×Ey±z%ai
with
respect
to
a.
i.
1)
.
So
taking
e
clockwise
with
respect
to
(
1.
1.
1)
Cutting
out
by
the
phone
xtyt
2-
=p
We
hav e
§
ydxtzdytxdz
=
tea
.
☐
